Alpha Decay
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(Avogadro number =6.023×1023 mo1−1)
- 1.725×104 s−1
- 1.535×104 s−1
- 1.035×104 s−1
- 1.235×104 s−1
Hydrogen bomb is based on the principle of
Nuclear fission
Nuclear fusion
Nuclear explosion
Chemical reaction
Why is binding energy important?
- 1.1 MeV
- 4.3 MeV
- 8.2 MeV
- 15.9 MeV
- N=nλ+N0e−λt
- N=N0e−λt
- N=nλ+(N0+nλ)e−λt
- N=nλ+(N0−nλ)e−λt
- q<3Q8
- q>3Q2
- q<8Q3
- q>3Q8
- M2=2 M1
- M2>2 M1
- M2<M1+10(mp+mn)
- M1< 10 (mn+mp)
- 10.8 MeV
- 10.9 MeV
- 11 MeV
- 11.1 MeV
- t12>T
- tavg=T
- t12=T
- tavg>T
[1amu c2=931.5 MeV].
- 2.28×104
- 6.3×105
- 3×106
- 1.8×05
11H1.007825 u21H2.014102 u63Li6.015123 u73Li7.016004 u15264Gd151.919803 u20682Pb205.974455 u31H3.016050 u41He4.002603 u7030Zn69.925325 u8234Se81.916709 u20983Bi208.980388 u21084Po209.982876 u
(1 u=932MeVc2)
A neutron having kinetic energy 12.5 eV collides with a hydrogen atom at rest. Neglect the difference in mass between the neutron and the hydrogen atom and assume that the neutron does not leave its line of motion. Find the possible kinetic energies of the neutron after the event.
21283 Bi can disintegrate either by emitting an α -particle or
by emitting a β− -particle. (a) Write the two equations showing the products of the decays. (b) The probabilities of disintegration by α - and β -decays are in the ratio 713. The overall half-life of 212 Bi is one hour. If 1 g of pure 212 Bi is taken at 12.00 noon, what will be the composition of this sample at 1 p.m. the same day?
- 79.0 eV
- 51.8 eV
- 49.2 eV
- 38.2 eV
- 1:√2
- 1:8
- 4:1
- 1:4
- 154 MeV
- 5411 MeV
- 5554 MeV
- 2711 MeV
Deffrentiate alpha decay, gamma decay, beta decay
A radioactive nucleus is being produced at a constant rate α per second. Its decay constant is λ If N0are the
number of nuclei at time t = 0, then maximum number of nuclei possible are
- 10 days and 40 days
- 5 days and 10 days
- 20 days and 5 days
- 20 days and 10 days
- 38 s
- 34 s
- 12 s
- 1 s
The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni= 1.5 × 1016 m−3. Is the material n-type or p-type?